Density Calculations Chemistry Tutorial

⚛ Density is defined as mass per unit volume:

density = mass/volume

⚛Density can be calculated using the mathematical formula (expression):

d = m ÷ V   or   ρ = m ÷ V
where d or ρ = density, m = mass, V = volume

⚛ The greater the density, the more mass per unit volume.

⚛ The unit of density derived from SI units is kilograms per cubic meter, kg/m³ or kg m^-3.

⚛ More commonly, densities are given in g/mL (g mL^-1) or g/cm³ (g cm^-3 or g/cc).

⚛ Density is a characteristic property of pure substances so density can help identify a particular pure substance.

⚛ Since different substances have different densities, density can also be used to separate one substance from another.

⚛ In general, solids are more dense than liquids which are more dense than gases:

density of solid > density of liquid > density of gas

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Density is defined as the amount of mass per unit volume.
You can determine, or measure, the density of solids, liquids and gases in the lab.

Density of Solids

The metallic elements aluminium and gold are both solids at room temperature and pressure. By weighing a known volume of metal we can determine the density of each metal.

Imagine 2 cubes. Each cube measures 1 cm × 1 cm × 1cm so that each cube has a volume of 1 cm³.
One cube is made out of gold and has a mass of 19.3 grams.
The other cube is made out of aluminium has a mass of only 2.7 grams.
Gold is said to be more dense than aluminium because, when we compare the same volume of the two different substances, this volume of gold has the greater mass.
We can also say that aluminium is less dense than gold because, for the same volume of the two substances, this volume of aluminium has less mass.

The density of each substance is given in units of mass per unit volume, for our cubes this will be mass in grams per cubic centimeter, that is, g/cm^-3 or g cm^-3.
Gold has a density of 19.3 grams per cubic centimeter, that is, 19.3 g/cm³ or 19.3 g cm^-3.
Aluminium has a density of 2.7 grams per cubic centimeter, that is, 2.7 g/cm³ or 2.7 g cm^-3.

You probably have some 1 cm³ cubes of different substances in your lab. You can use these cubes to determine the densities of some different solid substances like copper, iron, zinc etc.

Density of Liquids

If you want to measure the density of liquids in the lab, it is much easier to use units of mass per unit volume in millilitres.
You could use a 1.00 mL pipette to acquire a sample of the liquid and weigh this.
If you weigh 1.00 mL of liquid water it will have a mass of about 1 g so its density is about 1 g/mL or 1 g mL^-1.
If you weigh 1.00 mL of vegetable oil it might have a mass of about 0.8 g so its density is about 0.8 g/mL or 0.8 g ml^-1.
Liquid water is more dense than vegetable oil. So, if you pour water and vegetable oil into the same beaker they will not mix but they will separate out into 2 layers. You will find the layer of more dense water at the bottom of the beaker and the less dense vegetable oil layer lying above the water. If you are very careful, you can "pour off", or "decant", the vegetable oil leaving the water in the beaker.
This concept of separating out layers of liquids based on their density is very useful in chemistry.
Oil and petrol (gasoline) do not mix with water. They are both less dense than water so they will also float on top of water. You may have seen pictures of oil spills from ships at sea in which you can see the black oil floating on top of the water, and then washing up onto the shore. Because the oil floats on top of water, it is possible to contain the oil spill at sea before it washes ashore by building a floating wall around it then "scooping up" the oil from the surface.

Density of Gases

The density of gases can also be measured.
The density of the air you breath is about 0.001 g mL^-1.
The density of helium gas is about 0.00016 g mL^-1.
Helium gas is less dense than the air around you.
If you fill a balloon with helium gas you need to hold onto the string otherwise it will float up and away!
The density of carbon dioxide gas is about 0.0018 g mL^-1. Carbon dioxide gas is more dense than the air around you. If you filled a balloon with pure carbon dioxide gas it will not float up, it will sink down to the floor!
This property of carbon dioxide gas has been used to fight fires, firstly because carbon dioxide gas doesn't burn, and secondly because the carbon dioxide gas can form a floating barrier that blocks off the supply of oxygen gas so that the substance can no longer burn.

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The density of a some pure substances is given in the table below. Can you find some patterns in the data?

In general, metals are more dense than non-metals. ⁽¹⁾
For example, at 25°C the density of metallic lead is 11.4 g mL^-1 while the density of non-metallic sulfur is only 2.0 g mL^-1.
This means that the mass of 1 mL of lead would be 11.4 g while the mass of 1 mL of sulfur would be only 2.0 g.

In general, solids are more dense than liquids, and liquids are more dense than gases. ⁽²⁾
For example, at 25°C, solid sulfur has a density of 2.0 g mL^-1, liquid water has a density of 1.0 g mL^-1, and gaseous oxygen has a density of 0.0013 g mL^-1.

There are some exceptions to this.
For example, balsa wood can be used to make model aeroplanes and kites. It has a density of 0.12 g mL^-1 so solid balsa wood is much less dense than other solids. Indeed, balsa wood is even less dense than liquid water and will float on top of water.

Because the atoms (or molecules) in a solid generally pack together very tightly they occupy the smallest volume.
In a liquid, the same mass of atoms (or molecules) are packed less tightly together so they occupy a greater volume and the density of the substance therefore decreases.
In a gas, the same mass of atoms (or molecules) are more randomly distributed in a much greater volume, so, the density of a gas is much, much, less than the density of either the liquid or the solid.

In general, for any pure substance, the density of the solid is greater than the density of the liquid which is greater than the density of the gas.

density of solid > density of liquid > density of gas

There are some exceptions to this.
For example, you have probably seen ice (solid water) floating on top of liquid water. This means that the solid water (ice) MUST be less dense than the liquid water!

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Density is defined as mass per unit volume, that is:

• density = mass/unit volume
• density = mass ÷ volume

If we let d = density, m = mass, and V = volume

d = m/V
or
d = m ÷ V

The Greek letter rho, ρ, can also be used to represent density. So the mathematical expression for the relationship between a substance's mass and volume can also be given as

ρ = m/V
or
ρ = m ÷ V

If we measure mass in units of grams (g) and volume in units of millilitres (mL), then

mass (g) ÷ volume (mL) = density (g mL^-1)
density has units of grams per millilitre (g mL^-1)

If we measure mass in units of grams (g) and volume in units of cubic centimetres (cm³), then

mass (g) ÷ volume (cm³) = density (g cm^-3)
density has units of grams per cubic centimetre (g cm^-3)

Therefore we can calculate the density of any substance by dividing its mass by its volume: d = m ÷ V

You can look up the densities of many substances in tables, like the one above.
By re-arranging the mathematical expression d = m/V (or ρ = m/V) we can use the tabulated value of density to calculate:

(i) mass of substance (m) if you know its volume (V)

m = d × V

(ii) volume of substance (V) if you know its mass (m)

V = m ÷ d

Always check for consistency in units in your calculations:

• If density is given in g mL^-1 then the mass must be in grams and the volume in millilitres.
• If density is given in g cm^-3 then the mass must be in grams and the volume in cubic centimetres.

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Worked Examples: Density Calculations

Problem solving based on the StoPGoPS approach

1. Step 1: What is the question asking you to do?
2. Step 2: What information have you been given?
3. Step 3: What is the relationship between what you know and what you need to find out?
4. Step 4: Substitute the values into the mathematical expression for density and solve
5. Step 5: Check your answer
6. Step 6: State the answer to the question

Question 1. Calculate the density in g cm^-3 of a ruby which has a volume of 1.6 cm³ and a mass of 6.7 g.

Step 1: What is the question asking you to do?

Calculate density (d) in g cm^-3

Step 2: What information have you been given?

m = mass = 6.7 g

V = volume = 1.6 cm³

Step 3: What is the relationship between what you know and what you need to find out?

Write the mathematical expression for calculating density:

d (g cm^-3) = m (g) ÷ V (cm³)

Step 4: Substitute the values into the mathematical expression and solve for density:

d (g cm^-3) = m (g) ÷ V (cm³)

d = 6.7 g ÷ 1.6 cm³ = 4.2 g cm^-3

Step 5: Check your answer:

(i) Have you answered the question that was asked?
Yes, we have calculated density in g cm^-3 as asked.

(ii) Have you used the correct number of significant figures?
mass (6.7 ) has 2 significant figures
volume (1.6) has 2 significant figures
When multiplying or dividing, the number of significant figures in the result is the same as the least number of significant figures in the terms used, therefore we are justified in using 2 significant figures.
Our answer (4.2) has 2 significant figures.

(iii) Is your answer plausible?
Perform a rough calculation:
mass ≈ 8 g
volume ≈ 2 cm³
density ≈ 8 ÷ 2 = 4 g cm^-3
Since our "rough" calculation is about the same as the carefully calculated density we are reasonably confident that our answer is correct.

Step 6: State the answer to the question:

density = 4.2 g cm^-3

Question 2. Calculate the density in g mL^-1 of a liquid which has a volume of 28 mL and a mass of 26.4 g

Step 1: What is the question asking you to do?

Calculate density (d) in g mL^-1

Step 2: What information have you been given?

m = mass = 26.4 g

V = volume = 28 mL

Step 3: What is the relationship between what you know and what you need to find out?

Write the mathematical expression for calculating density:

d (g mL^-1) = m (g) ÷ V (mL)

Step 4: Substitute the values into the mathematical expression and solve for density:

d (g mL^-1) = m (g) ÷ V (mL)

d (g mL^-1) = 26.4 (g) ÷ 28 (mL) = 0.94 g mL^-1

Step 5: Check your answer

(i) Have answered the question that was asked?
Yes, we have calculated the density in g mL^-1 as asked.

(ii) Have you used the correct number of significant figures?
mass (26.4) has 3 significant figures
volume (28) has 2 significant figures
When multiplying or dividing, the number of significant figures in the result is the same as the least number of significant figures in the terms used, therefore we are justified in using 2 significant figures.
Our answer (0.94) has 2 significant figures.

(iii) Is your answer plausible?
Perform a "rough" calculation:
mass ≈ 25 g
volume ≈ 25 mL
density ≈ 25 g ÷ 25 mL = 1 g mL^-1
Since our answer for the "rough" calculation is about the same as our carefully calculated answer we are reasonably confident that our answer is correct.

Step 6: State the answer to the question:

density = 0.94 g mL^-1

Question 3. Beeswax has a density of 0.96 g cm^-3 at 25°C and 1 atm pressure.
Calculate the mass in grams of 5.0 cm³ of beeswax at 25°C and 1 atm pressure.

Step 1: What is the question asking you to do?

Calculate mass (m) in grams (g).

Step 2: What information have you been given?

d = density = 0.96 g cm^-3 (at 25°C and 1 atm)

V = volume = 5.0 cm³ (at 25°C and 1 atm)

Step 3: What is the relationship between what you know and what you need to find out?

Write the mathematical expression for calculating density:

d (g cm^-3) = m (g) ÷ V (cm³)

Step 4: Substitute the values into the mathematical expression and solve for mass (m):

d (g cm^-3) = m (g) ÷ V (cm³)

0.96 g cm^-3 = m (g) ÷ 5.0 cm³

Multiply both sides of the equation by 5.0 cm³:

0.96 g cm^-3 × 5.0 (cm³) = 5 (cm³) × m (g) ÷ (5 cm³)

4.8 g = m (g)

Step 5: Check your answer

(i) Have answered the question that was asked?
Yes, we have calculated the mass in grams as asked.

(ii) Have you used the correct number of significant figures?
density (0.96 ) has 2 significant figures
volume (5.0) has 2 significant figures
When multiplying or dividing, the number of significant figures in the result is the same as the least number of significant figures in the terms used, therefore we are justified in using 2 significant figures.
Our answer (4.8) has 2 significant figures.

(iii) Is your answer plausible?
Density is about 1 g cm^-1, that is, a mass of 1 g has a volume of 1 cm^-1.
A volume of 5 cm^-1 will therefore have a mass of 5 × 1 = 5 cm³
Since this "rough" calculation of mass agrees with our carefully calculated value for mass we are reasonably confident that our answer is correct.

Step 6: State the answer to the question:

mass = 4.8 g

Question 4. Milk has a density of 1.03 g mL^-1 at 25°C and 1 atm pressure.
Calculate the mass in grams of 1 L of milk at 25°C and 1 atm.

Step 1: What is the question asking you to do?

Calculate the mass (m) in grams (g)

Step 2: What information have you been given?

d = density = 1.03 g mL^-1 (at 25°C and 1 atm)

V = volume = 1 L (at 25°C and 1 atm)

Convert volume in L to volume in mL so the units for density (g mL^-1) will be consistent:

V = 1 L × 1000 mL L^-1 = 1000 mL

Step 3: What is the relationship between what you know and what you need to find out?

Write the mathematical expression for calculating density:

d (g mL^-1) = m (g) ÷ V (mL)

Step 4: Substitute the values into the mathematical expression and solve for mass (m):

d (g mL^-1) = m (g) ÷ V (mL)

1.03 g mL^-1 = m (g) ÷ 1000 mL

Multiple both sides of the equation by 1000 mL:

1.03 g mL^-1 × 1000 mL = 1000 mL × m (g) ÷ 1000 mL

1030 g = m (g)

Step 5: Check your answer

(i) Have answered the question that was asked?
Yes, we have calculated the mass in grams as asked.

(ii) Have you used the correct number of significant figures?
density (1.03 ) has 3 significant figures
volume (1) has 1 significant figure
When multiplying or dividing, the number of significant figures in the result is the same as the least number of significant figures in the terms used, therefore we are justified in using 1 significant figure.
Our answer (1030) has 4 significant figures!
Since we are only justified in using 1 significant figure, we could give the answer as 1000 g, but the best answer would be to use scientific notation and write 1 × 10³ g

(iii) Is your answer plausible?
Perform a "rough" calculation:
density ≈ 1 g mL^-1
1 mL of milk has a mass of about 1 g
1 L (1000 mL) of milk has a mass of about 1000 × 1 = 1000 g
Since this rough calculation is close to the carefully calculated value for mass we are reasonably confident that our answer is correct.

Step 6: State the answer to the question:

mass = 1000 g = 1 × 10³ g

Question 5. At 25°C and 1 atm pressure, diamond has a density of 3.5 g cm^-3.
Calculate the volume in cm³ of 0.50 g of diamond at 25°C and 1 atm pressure.

Step 1: What is the question asking you to do?

Calculate the volume (V) in cm³

Step 2: What information have you been given?

d = density = 3.5 g cm^-3 (at 25°C and 1 atm)

m = mass = 0.50 g (at 25°C and 1 atm)

Step 3: What is the relationship between what you know and what you need to find out?

Write the mathematical expression for calculating density:

d (g cm^-3) = m (g) ÷ V (cm³)

Step 4: Substitute the values into the mathematical expression and solve for density:

d (g cm^-3) = m (g) ÷ V (cm³)

3.5 g cm^-3 = 0.50 g ÷ V (cm³)

Multiply both sides of the equation by V (cm³):

3.5 g cm^-3 × V (cm³) = V(cm³) × 0.50 g ÷ V (cm³)
3.5 g cm^-3 × V (cm³) = 0.50 g

Divide both sides of the equation by 3.5 g cm^-3:

[3.5 g cm^-3 × V (cm³)] ÷ 3.5 g cm^-3 = 0.50 g ÷ 3.5 g cm^-3
V (cm³) = 0.50 ÷ 3.5 (cm^-3) = 0.14 cm³

Step 5: Check your answer

(i) Have you answered the question that was asked?
Yes, we have calculated the volume in cm³.

(ii) Have you used the correct number of significant figures?
density (3.5) has 2 significant figures
volume (0.50) has 2 significant figures
When multiplying or dividing, the number of significant figures in the result is the same as the least number of significant figures in the terms used, therefore we are justified in using 2 significant figures.
Our answer (0.14) has 2 significant figures

(iii) Is your answer plausible?
Perform a "rough" calculation:
density ≈ 5 g cm^-3
1 cm^-3 of diamond has a mass of about 5 g.
The question gives us 0.50 g for the mass which is ¹/₁₀^th the mass of 1 cm^-3 of diamond
So the volume of 0.50 g of diamond will ¹/₁₀^th the volume that is
V = ¹/₁₀ × 1 cm³ = 0.1 cm³
Since this "rough" answer is about the same as our carefully calculated answer we are reasonably confident that our answer is correct.

Step 6: State the answer to the question:

volume = 0.14 cm³

Question 6. At 25°C and 1 atm pressure, ethylene glycol has a density of 1.11 g mL^-1.
Calculate the volume in mL of 0.025 kg of ethylene glycol at 25°C and 1 atm pressure.

Step 1: What is the question asking you to do?

Calculate the volume (V) in mL

Step 2: What information have you been given?

d = density = 1.11 g mL^-1 (at 25°C and 1 atm pressure)

m = mass = 0.025 kg (at 25°C and 1 atm pressure)

Convert mass in kg to mass in g in order to be consistent with units of density (g mL^-1):

m = 0.025 kg × 1000 g kg^-1 = 25 g

Step 3: What is the relationship between what you know and what you need to find out?

Write the mathematical expression for calculating density:

d (g mL^-1) = m (g) ÷ V (mL)

Step 4: Substitute the values into the mathematical expression and solve for density:

d (g mL^-1) = m (g) ÷ V (mL)

1.11 (g mL^-1) = 25 (g) ÷ V (mL)

Multiply both sides of the equation by V (mL):

V (mL) × 1.11 (g mL^-1) = V (mL) × 25 (g) ÷ V (mL)
V (mL) × 1.11 (g mL^-1) = 25 (g)

Divide both sides of the equation by 1.11 (g mL^-1):

[V (mL) × 1.11 (g mL^-1)] ÷ 1.11 (g mL^-1) = 25 (g) ÷ 1.11 (g mL^-1)
V (mL) = 25 ÷ 1.11 (mL^-1) = 23 mL

Step 5: Check your answer

(i) Have you answered the question that was asked?
Yes, we have calculated the volume in mL as asked.

(ii) Have you used the correct number of significant figures?
mass (0.025) has 2 significant figures
density (1.11) has 3 significant figures
When multiplying or dividing, the number of significant figures in the result is the same as the least number of significant figures in the terms used, therefore we are justified in using 2 significant figures.
Our answer (25) has 2 significant figures

(iii) Is your answer plausible?
Work backwards, that is, use the mass given and our value for volume to see if we get the same value for density as given in the question.
m = 0.025 kg = 25 g
V = 23 mL
d = 25 ÷ 23 = 1.1 g mL^-1
Since this value for density agrees with that given in the question we are reasonably confident that our answer is correct.

Step 6: State the answer to the question:

volume = 23 mL

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A student weighed a 25.00 mL sample of an unknown liquid and found its mass to be 18.158 g at 25°C and 1 atmosphere pressure.

The densities of some liquid compounds at 25°C and 1 atmosphere pressure are given in the table below:

Identify the name of the liquid.

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